Let m and n be the two positive integers greater than 1 . If `underset( alpha rarr 0 ) ( "lim")((e^(cos(alpha^(n)))-e)/(alpha^(m))) = -((e )/( 2))` then the value of `( m )/( n )` is

Answer

To solve the limit problem, we start with the expression given:

\[ \lim_{\alpha \to 0} \frac{e^{\cos(\alpha^n)} - e}{\alpha^m} = -\frac{e}{2} \]

### Step 1: Evaluate \(\cos(\alpha^n)\) as \(\alpha \to 0\)

As \(\alpha\) approaches 0, \(\alpha^n\) also approaches 0. Therefore, we can use the Taylor series expansion for \(\cos(x)\) around 0:

\[ \cos(x) \approx 1 - \frac{x^2}{2} \quad \text{for small } x \]

Substituting \(x = \alpha^n\):

\[ \cos(\alpha^n) \approx 1 - \frac{(\alpha^n)^2}{2} = 1 - \frac{\alpha^{2n}}{2} \]

### Step 2: Substitute \(\cos(\alpha^n)\) into the limit

Now we substitute this approximation into the limit:

\[ e^{\cos(\alpha^n)} \approx e^{1 - \frac{\alpha^{2n}}{2}} = e \cdot e^{-\frac{\alpha^{2n}}{2}} \approx e \left(1 - \frac{\alpha^{2n}}{2}\right) \quad \text{(using } e^x \approx 1 + x \text{ for small } x\text{)} \]

Thus, we have:

\[ e^{\cos(\alpha^n)} - e \approx e \left(1 - \frac{\alpha^{2n}}{2}\right) - e = -\frac{e \alpha^{2n}}{2} \]

### Step 3: Substitute into the limit expression

Now we substitute this back into our limit:

\[ \lim_{\alpha \to 0} \frac{-\frac{e \alpha^{2n}}{2}}{\alpha^m} \]

This simplifies to:

\[ \lim_{\alpha \to 0} -\frac{e}{2} \cdot \frac{\alpha^{2n}}{\alpha^m} = -\frac{e}{2} \cdot \lim_{\alpha \to 0} \alpha^{2n - m} \]

### Step 4: Determine the condition for the limit to exist

For the limit to exist and equal \(-\frac{e}{2}\), the exponent of \(\alpha\) must be zero. Therefore, we set:

\[ 2n - m = 0 \implies m = 2n \]

### Step 5: Find the ratio \( \frac{m}{n} \)

Now, we can find the ratio:

\[ \frac{m}{n} = \frac{2n}{n} = 2 \]

### Final Answer

Thus, the value of \(\frac{m}{n}\) is:

\[ \boxed{2} \]